Derivation of Euler-Lagrange equations

One degree of freedom

For any curve $α : I \mapsto C$ , where $C$ is the configuration space of Lagrangian Mechanics, we can create

\tilde{α} = (α, α^{'}) : I ⟼ T C

(is nothing but the prolongation of a section of the trivial bundle $R \times C \to R$ ).

Then, the action for that curve is the integral of the composition

L \circ \tilde{α} : I ⟼ R

where $L : T C \mapsto R$ is a Lagrangian. That is

S_{α} = \int_{t 0}^{t_{1}} L \circ \tilde{α} d t

with fixed $\tilde{α} (t_{i})$ for every $α$ .

(in the context of the variational bicomplex, the Lagrangian is a differential form $λ = L d t$ of type $(1, 0)$ and the action is $\int_{T} {\tilde{α}}^{*} (λ)$ , see Anderson_1992 page 9).

The trick to find $α$ such that $S_{α}$ is "critical" is what follows. Consider that the whole $\tilde{α}$ lie inside a local chart (if not, I think that it could be done in a partition of unity, the same that we would have for the integration process). In local coordinates, let $\tilde{α} (t) = (q (t), q^{'} (t))$ , with $q (t) \in R^{n}$ .

We can choose any vector field over ${q (t)}_{t \in I}$ , for example $V (t)$ , and a little real number $ϵ$ to construct a new curve $β_{ϵ}$ that in local coordinates looks like

(q (t) + ϵ V (t))

This curve gives us a $\tilde{β_{ϵ}}$ (the prolongation of $β_{ϵ}$ ), which in local coordinates is written

(q (t) + ϵ V (t), q^{'} (t) + ϵ V^{'} (t))

(By the way, we are deriving using the local chart connection).

If we consider the smooth function

ϵ \to S_{β_{ϵ}} = \int_{t_{0}}^{t_{1}} L (q (t) + ϵ V (t), q^{'} (t) + ϵ V^{'} (t)) d t,

its derivative respect to $ϵ$ should be null for $ϵ = 0$ in order to $α$ be a extrema of this functional. So, let's go. Consider that the specified local char induces in $T C$ coordinates $(q, v)$ , then:

\int_{t_{0}}^{t_{1}} \frac{\partial L}{\partial q} V + \frac{\partial L}{\partial v} V^{'} d t = 0

Integrating the second term by parts:

\int_{t_{0}}^{t_{1}} \frac{\partial L}{\partial q} V d t + {[\frac{\partial L}{\partial v} V]}_{t_{0}}^{t_{1}} - \int_{t_{0}}^{t_{1}} \frac{d}{d t} (\frac{\partial L}{\partial v}) V d t = 0

And since the end points are fixed

\int_{t_{0}}^{t_{1}} \frac{\partial L}{\partial q} V - \frac{d}{d t} (\frac{\partial L}{\partial v}) V d t = 0

And since this is valid for every $V$ we get, finally:

\frac{\partial L}{\partial q} (α (t), α^{'} (t)) = \frac{d}{d t} (\frac{\partial L}{\partial v} (α (t), α^{'} (t)))

or more compactly

\frac{\partial L}{\partial q} - \frac{d}{d t} \frac{\partial L}{\partial v} = 0

It is a straightforward calculation to check that Euler-Lagrange equations "behave in a good way" for change of variables.

This is generalized by the Euler operator: see the end of variational derivative#Some facts.

Discrete approach

xournal_154

Infinite degree of freedom: fields

With the notation of the jet bundle $J^{1} (E)$ :

\frac{\partial L}{\partial u^{α}} - \frac{d}{d x^{i}} \frac{\partial L}{\partial u_{i}^{α}} = 0

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Author of the notes: Antonio J. Pan-Collantes

antonio.pan@uca.es

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